RGB图像的PCA [英] PCA of RGB Image

问题描述

我试图弄清楚如何使用PCA在python中解相关RGB图像. 我正在使用O'Reilly Computer视觉书籍中的代码:

I'm trying to figure out how to use PCA to decorrelate an RGB image in python. I'm using the code found in the O'Reilly Computer vision book:

from PIL import Image
from numpy import *

def pca(X):
  # Principal Component Analysis
  # input: X, matrix with training data as flattened arrays in rows
  # return: projection matrix (with important dimensions first),
  # variance and mean

  #get dimensions
  num_data,dim = X.shape

  #center data
  mean_X = X.mean(axis=0)
  for i in range(num_data):
      X[i] -= mean_X

  if dim>100:
      print 'PCA - compact trick used'
      M = dot(X,X.T) #covariance matrix
      e,EV = linalg.eigh(M) #eigenvalues and eigenvectors
      tmp = dot(X.T,EV).T #this is the compact trick
      V = tmp[::-1] #reverse since last eigenvectors are the ones we want
      S = sqrt(e)[::-1] #reverse since eigenvalues are in increasing order
  else:
      print 'PCA - SVD used'
      U,S,V = linalg.svd(X)
      V = V[:num_data] #only makes sense to return the first num_data

   #return the projection matrix, the variance and the mean
   return V,S,mean_X

我知道我需要弄平图像，但是形状是512x512x3. 3的维度会影响我的结果吗?我该如何截断? 我如何找到保留多少信息的定量数量?

I know I need to flatten my image, but the shape is 512x512x3. Will the dimension of 3 throw off my result? How do I truncate this? How do I find a quantitative number of how much information is retained?

推荐答案

如果有三个波段(RGB图像就是这种情况)，则需要像这样重塑图像

If there are three bands (which is the case for an RGB image), you need to reshape your image like

X = X.reshape(-1, 3)

在512x512图像的情况下，新的X将具有形状(262144, 3). 3维将不会影响您的结果；该维度表示图像数据空间中的特征. X的每一行都是一个样本/观测值，每一列代表一个变量/特征.

In your case of a 512x512 image, the new X will have shape (262144, 3). The dimension of 3 will not throw off your result; that dimension represents the features in the image data space. Each row of X is a sample/observation and each column represents a variable/feature.

图像中的总方差等于np.sum(S)，这是特征值的总和.您保留的方差量将取决于您保留的特征值/特征向量.因此，如果仅保留第一个特征值/特征向量，则保留的图像方差分数将等于

The total amount of variance in the image is equal to np.sum(S), which is the sum of eigenvalues. The amount of variance you retain will depend on which eigenvalues/eigenvectors you retain. So if you only keep the first eigenvalue/eigenvector, then the fraction of image variance you retain will be equal to

f = S[0] / np.sum(S)

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